(n+3)(n-6)=10

Solution for (n+3)(n-6)=10 equation:

(n+3)(n-6)=10

We move all terms to the left:

(n+3)(n-6)-(10)=0

We multiply parentheses ..

(+n^2-6n+3n-18)-10=0

We get rid of parentheses

n^2-6n+3n-18-10=0

We add all the numbers together, and all the variables

n^2-3n-28=0

a = 1; b = -3; c = -28;
Δ = b2-4ac
Δ = -32-4·1·(-28)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*1}=\frac{-8}{2}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*1}=\frac{14}{2}
=7
$